0
$\begingroup$

Let $X$ be a random variable with mean, $E(X)=\mu$ and variance, $E(X-\mu)^2=\sigma^2$. Then Chebyshev's inequality asserts that $$ P\{|X-\mu|\geq k\sigma\} \leq \frac{1}{k^2} $$

Using this inequality, I have to show that $$e^{k+1} \geq k^2 \,\,\text{for}\,\, k>1$$ It is clear that if I can show that $$P\{|X-\mu|\geq k\sigma\}\geq \frac{1}{e^{k+1}}$$ for $k>1$, and any $\mu \in \mathbb{R}$ and any $\sigma>0$, then we are done. Thanks in advance.

Source : Rohatgi, Saleh-p.$98$-problem $6$.

$\endgroup$
2
$\begingroup$

Consider a random variable $X$ that follows the exponential distribution $\mathsf{Exp}(1)$. We have $$ \mathsf{E}(X) = 1, \quad\mathsf{Var}(X) = 1 $$ and for $x > 0$, $$ \Pr(X \geq x) = e^{-x} $$ implying $\Pr(X \geq k + 1) = 1/ e^{k+1}$. In addition, by Chebyshev's bound, we have $$ \Pr(X \geq 1 + k) \leq 1/k^2 $$ Therefore, $$ 1 / e^{k+1} \leq 1/k^2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.